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6.4 Answer

6.4

1.

(a) $f_{x}(x,y) = 3x^2 y^4, f_{y}(x,y) = 4x^3 y^3$より

$\displaystyle df(x,y)$ $\displaystyle =$ $\displaystyle 3x^2 y^4 dx + 4x^3 y^3 dy$  
$\displaystyle \nabla f(x,y)$ $\displaystyle =$ $\displaystyle (3x^2 y^4, 4x^3 y^3)$  

The tangent plane goes through the point corresponding to $(1,1)$ is
$\displaystyle z$ $\displaystyle =$ $\displaystyle f(1,1) + \nabla f(1,1) \cdot (x - 1,y-1)$  
  $\displaystyle =$ $\displaystyle 1 + 3(x-1) + 4(y-1) = 3x + 4y -6$  

The normal line goes through the point corresponds to $(1,1)$ is

$\displaystyle (x,y,z) - (1,1,1) = t(3,4,-1)$

or，

$\displaystyle t = \frac{x-1}{3} = \frac{y-1}{4} = \frac{z -1}{-1}$

(b)

$f_{x}(x,y) = 3x^2 y + 2xy^4, f_{y}(x,y) = x^3 + 4x^2 y^3$ implies

$\displaystyle df(x,y)$ $\displaystyle =$ $\displaystyle (3x^2y + 2xy^4) dx + (x^3 + 4x^2 y^3) dy$  
$\displaystyle \nabla f(x,y)$ $\displaystyle =$ $\displaystyle (3x^2y + 2xy^4, x^3 + 4x^2 y^3)$  

The tangent plane goes through the point corresponds to $(1,1)$ is
$\displaystyle z$ $\displaystyle =$ $\displaystyle f(1,1) + \nabla f(1,1) \cdot (x - 1,y-1)$  
  $\displaystyle =$ $\displaystyle 2 + (5,5) \cdot (x-1,y-1) = 2 + 5x - 5 + 5y - 5$  
  $\displaystyle =$ $\displaystyle 5x + 5y - 8$  

The normal line goes through the point corresponds to $(1,1)$ is

$\displaystyle (x,y,z) - (1,1,2) = t(5,5,-1)$

or

$\displaystyle t = \frac{x-1}{5} = \frac{y-1}{5} = \frac{z -2}{-1}$

(c)

$z_{x} = 2xye^{2x} + 2x^2 y e^{2x}, z_{y} = x^2 e^{2x}$より

$\displaystyle dz$ $\displaystyle =$ $\displaystyle (2xye^{2x} + 2x^2 y e^{2x}) dx + x^2 e^{2x}dy$  
$\displaystyle \nabla f(x,y)$ $\displaystyle =$ $\displaystyle (2xye^{2x} + 2x^2 y e^{2x}, x^2 e^{2x})$  

THe tangent plane goes through the point corresponds to 点$(1,1)$ is
$\displaystyle z$ $\displaystyle =$ $\displaystyle e^2 + (2e^2 + 2e^2) \cdot (x - 1,y-1)$  
  $\displaystyle =$ $\displaystyle e^2 + 4e^2(x-1) + e^2(y-1)$  

The normal line goes through the point corresponds to $(1,1)$ is

$\displaystyle (x,y,z) - (1,1,e^2) = t(4e^2,e^2,-1)$

or

$\displaystyle t = \frac{x-1}{4e^2} = \frac{y-1}{e^2} = \frac{z - e^2}{-1}$

2.

(a) Consider $f(x,y) = \sqrt{x}\sqrt[4]{y}$. Then $f(121,16) = \sqrt{121}\sqrt[4]{16} = 22$．Then the seeking vlaue is $f(125,17) = f(121+4,16+1)$．Now let $\Delta x = 4, \Delta y = 1$. Then

$\displaystyle f(121+4,16+1)$ $\displaystyle =$ $\displaystyle f(121,16) + \Delta f(121,16)$  
  $\displaystyle \approx$ $\displaystyle f(121,16) + df(121,16)$  


$\displaystyle df$ $\displaystyle =$ $\displaystyle f_{x}dx + f_{y}dy$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{4}}dx + \frac{1}{4}x^{\frac{1}{2}}y^{-\frac{3}{4}}dy$  

$\Delta x = dx, \ \Delta y = dy$ Then

$\displaystyle df(121,16) = \frac{4}{11} + \frac{11}{32} \approx 0.7$

Therefore， $f(125,17) \approx 22.7$

(b) Consider $f(x,y) = \sin{x}\cos{y}$. Then $f(\pi,\frac{\pi}{3}) = \sin{\pi}\cos{\frac{\pi}{3}} = 0$．The seeking value is $f(\frac{6\pi}{7},\frac{\pi}{3}) = f(\pi - \frac{\pi}{7},\frac{\pi}{3})$．Now let $\Delta x = -\frac{\pi}{7}, \Delta y = 0$. Then

$\displaystyle f(\frac{6\pi}{7},\frac{\pi}{3})$ $\displaystyle =$ $\displaystyle f(\pi,\frac{\pi}{3}) + \Delta f(\pi,\frac{\pi}{3})$  
  $\displaystyle \approx$ $\displaystyle f(\pi,\frac{\pi}{3}) + df(\pi,\frac{\pi}{3})$  


$\displaystyle df$ $\displaystyle =$ $\displaystyle f_{x}dx + f_{y}dy$  
  $\displaystyle =$ $\displaystyle \cos{x}\cos{y}dx - \sin{x}\sin{y}dy$  

$\Delta x = dx, \ \Delta y = dy$ Thus,

$\displaystyle df(\pi,\frac{\pi}{3}) = \cos{\pi}\cos{\frac{\pi}{3}}dx - \sin{\pi}\sin{\frac{\pi}{3}}dy =\frac{\pi}{14} \approx 0.22$

Therefore， $f(\frac{6\pi}{7},\frac{\pi}{3}) \approx 0.22$

next up previous index
Next: Gradient and directional derivatives Up: PARTIAL DIFFERENTIATION Previous: Total differential   索引